/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Let \(f:[a, b] \rightarrow \math... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 42

Let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable and \(\phi:[m(f), M(f)] \rightarrow \mathbb{R}\) be continuous. Show that \(\phi \circ f:[a, b] \rightarrow \mathbb{R}\) is integrable. (Hint: Given \(\epsilon>0\), find \(\delta>0\) using the uniform continuity of \(\phi\). There is a partition \(P\) of \([a, b]\) such that \(U(P, f)-L(P, f)<\delta^{2}\). Divide the sum in \(U(P, f)-L(P, f)\) into two classes depending on whether \(M_{i}(f)-m_{i}(f)\) is less than \(\delta\), or greater than or equal to \(\delta\). Use the Riemann condition for \(\phi \circ f\).)

Short Answer

Expert verified
In summary, since $f$ is integrable and $\phi$ is continuous and uniformly continuous over the closed bounded interval $[m(f), M(f)]$, we can find a partition $P$ of $[a, b]$ that satisfies the Riemann condition for $\phi \circ f$. We divided the sum in $U(P, f) - L(P, f)$ into two parts depending on whether $M_i(f) - m_i(f) < \delta$ or $M_i(f) - m_i(f) \ge \delta$ and estimated the difference $U(P, \phi \circ f) - L(P, \phi \circ f)$. As a result, we found that $\phi \circ f$ is integrable over $[a, b]$.

Step by step solution

01

Find epsilon1 and delta using uniform continuity of phi

Since phi is uniformly continuous on [m(f), M(f)], given 蠁蔚鈧� > 0, there exists 未 > 0 such that for all x, y in [m(f), M(f)], if |x - y| < 未, then |蠁(x) - 蠁(y)| < 蠁蔚鈧�. Choose 蔚鈧� > 0 such that 蔚鈧� < 蔚/2 and find an appropriate 未 > 0.
02

Find a partition P for f such that the difference is less than delta square

Now let's find a proper partition P of [a, b] such that U(P, f) - L(P, f) < 未虏.
03

Divide the sum in U(P, f) - L(P, f) into two classes

As suggested in the hint, divide the sum in U(P, f) - L(P, f) into two parts depending on whether M岬�(f) - m岬�(f) < 未, or M岬�(f) - m岬�(f) 鈮� 未: 1. Sum for all i for which M岬�(f) - m岬�(f) < 未. 2. Sum for all i for which M岬�(f) - m岬�(f) 鈮� 未.
04

Apply the Riemann condition for phi鈭榝

Now let's estimate the difference U(P, 蠁鈭榝) - L(P, 蠁鈭榝) for the partition P. 1. For the first part, we know that if M岬�(f) - m岬�(f) < 未, then |蠁(M岬�(f)) - 蠁(m岬�(f))| < 蠁蔚鈧� since M岬�(f), m岬�(f) are in [m(f), M(f)] and |M岬�(f) - m岬�(f)| < 未. So in this case, for the difference U(P, 蠁鈭榝) - L(P, 蠁鈭榝) in part 1 would be at most 蔚鈧�. 2. For the second part, the total contribution is less than 未虏, by virtue of the selected partition P. Thus, U(P, 蠁鈭榝) - L(P, 蠁鈭榝) < 蔚鈧� + 未虏 < 蔚, as 蔚鈧� < 蔚/2 and 未虏 < 蔚/2. Hence, the required inequality is satisfied, and we can conclude that 蠁鈭榝 is integrable over [a, b], as we have found a partition P for which the Riemann condition is satisfied.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Riemann Condition
The Riemann condition is a crucial concept in understanding the integrability of functions. It states that a function is Riemann integrable on an interval if, for every positive number \( \(\epsilon\) \) you choose, there exists a partition of the interval such that the upper sum \( U(P, f) \) and the lower sum \( L(P, f) \) differ by less than \( \(\epsilon\) \).

In layman terms, if we can create a combination of rectangles that approximate the area under the curve of a function, and the difference between the estimates made with the tops and bottoms of these rectangles can be made arbitrarily small, then the function passes the Riemann condition. Thus, the Riemann condition serves as a test for determining if a function can be measured precisely in this manner, or in other words, if it can be integrated.
The Role of Uniform Continuity
Uniform continuity is a stronger form of continuity. A function is uniformly continuous on an interval if, no matter where you are on the interval, you can predict the function's output within a certain range by making sure your inputs are close enough together. Unlike regular continuity, which can vary in precision from one point to another, uniform continuity guarantees precision across the entire interval.

Uniform continuity comes into play with integrability when a function is composed with a continuous function, as in the problem \( \(\phi \circ f\) \). The reason it's essential here is that uniform continuity allows us to control the output of \( \(\phi\) \) by limiting the variation in \( \(f\) \) (since \( \(f\) \) is integrable). Consequently, this assures that the composite function \( \(\phi \circ f\) \) retains a level of predictability required for integrability.
Calculating the Definite Integral
A definite integral is a fundamental concept in calculus representing the area under the curve of a function over an interval. Formally, it's the limit of the sums of areas of rectangles (as mentioned under the Riemann condition) as they get infinitely thin and infinitely numerous.

The definite integral not only measures area but also accumulates quantities | think of it as a continuous sum over an interval. The process of calculating a definite integral is known as integration. Fundamentally, if a function is Riemann integrable over an interval, it means the function's definite integral exists over that interval, and we can calculate the 'total value' the function accumulates between two points.
Partitioning an Interval
Partitioning an interval is about dividing an interval into subintervals, which is central to understanding how Riemann sums work. Essentially, you're breaking an interval into smaller pieces to analyze a function's behavior more closely and sum up areas under the curve with better precision.

In the context of integrability, partitioning allows us to create upper and lower sums | by looking at the maximum and minimum values that a function can take on each subinterval. The smaller and more numerous the partitions, the closer these upper and lower sums can get to each other, enabling us to meet the Riemann condition for integrability by guaranteeing that our approximations of the function's integral become as accurate as necessary.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

2Let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable. Define \(G:[a, b] \rightarrow \mathbb{R}\) by $$ G(x):=\int_{x}^{b} f(t) d t . $$ Show that \(G\) is continuous on \([a, b] .\) Further, show that if \(f\) is continuous at \(c \in[a, b]\), then \(G\) is differentiable at \(c\) and \(G^{\prime}(c)=-f(c)\). (Hint: Propositions \(6.7,6.20\), and \(6.21\).)

Let \(D\) be a bounded subset of \(\mathbb{R}\). Let \(1_{D}: D \rightarrow \mathbb{R}\) be defined by \(1_{D}(x):=1\) for all \(x \in D\). Prove the following statements: (i) \(1_{D}\) is integrable if and only if \(\partial D\) is of content zero. [Note: If \(1_{D}\) is integrable, then \(\int_{D} 1_{D}(x) d x\) is called the length of the set \(D .]\) (ii) The length of \(D\) is zero if and only if \(D\) is of content zero. (iii) If \(f: D \rightarrow \mathbb{R}\) is a bounded function and \(D_{0} \subseteq D\) is such that \(\partial D\) is of content zero, then \(f\) is integrable on \(D_{0}\).

Let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable, and \(g:[a, b] \rightarrow \mathbb{R}\) be a bounded function such that the set \(\\{x \in[a, b]: g(x) \neq f(x)\\}\) is of content zero. Show that \(g\) is integrable and $$ \int_{a}^{b} g(x) d x=\int_{a}^{b} f(x) d x $$ (Compare Proposition 6.12.)

(Leibniz's Rule for Integrals) Let \(f\) be a continuous function on \([a, b]\) and \(u, v\) be differentiable functions on \([c, d] .\) If the ranges of \(u\) and \(v\) are contained in \([a, b]\), prove that $$ \frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t=\left[f(v(x)) \frac{d v}{d x}-f(u(x)) \frac{d u}{d x}\right] $$

Let \(f:[a, b] \rightarrow \mathbb{R}\) be a function. Show that \(f\) is integrable if (i) \(r f\) is integrable for some nonzero \(r \in \mathbb{R}\), or (ii) if \(f\) is bounded, \(f(x) \neq 0\) for all \(x \in[a, b]\), and \(1 / f\) is integrable.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.